Research and Communications in Mathematics and Mathematical Sciences Vol. 5, Issue 1, 2015, Pages 47-60 ISSN 2319-6939 Published Online on May 08, 2015  2015 Jyoti Academic Press http://jyotiacademicpress.net

INEQUALITIES ON THE LINE SEGMENT

ZLATKO PAVIĆ

Mechanical Engineering Faculty in Slavonski Brod University of Osijek Trg Ivane Brlić Mažuranić 2 35000 Slavonski Brod Croatia e-mail: [email protected]

Abstract

The article deals with generalizations of the inequalities for convex functions on the line segment. The Jensen and the Hermite-Hadamard inequalities are included in the study. Some improvements of the Hermite-Hadamard inequality are obtained and applied to mathematical .

1. Introduction

Let X be a real linear space. A linear combination αa + βb of points a, b ∈ X and coefficients α, β ∈ R is affine if α + β = 1. A set S ⊆ X is affine if it contains all binomial affine combinations of its points. A h : S → R is affine if the equality

h()αa + βb = αh(a) + βh(b), (1) holds for every binomial affine combination αa + βb of the affine set S.

2010 Mathematics Subject Classification: 26A51, 26D15. Keywords and phrases: convex function, convex combination, barycenter. Communicated by Atid Kangtunyakarn. Received January 30, 2015 48 ZLATKO PAVIĆ

Convex combinations and sets are introduced by restricting to affine combinations with nonnegative coefficients. A function h : S → R is convex if the inequality

f ()αa + βb ≤ αf (a) + βf (b), (2) holds for every binomial convex combination αa + βb of the convex set S.

Using mathematical induction, the above concept can be extended to n-membered affine or convex combinations.

In this paper, we use the real line X = R. Besides convex and affine combinations, we will use barycenters of the sets of real numbers. If µ is a positive measure on R, and if S ⊆ R is a measurable set such that µ()S > 0, then the integral point

1 c = x dµ, (3) µ()S ∫S is called the barycenter of the set S respecting the measure µ, or just the set barycenter. The barycenter c belongs to the convex hull of the set S, as the smallest convex set containing S. Given the measurable set S of positive measure, every affine function h : R → R satisfies the equality

 1  1 h  x dµ = h()x dµ. (4)  µ()S ∫S  µ()S ∫S

For the purpose of the paper, the set S will be used as an interval or a union of intervals.

2. The Jensen and the Hermite-Hadamard Inequalities

Through the paper, we will use a bounded interval of real numbers with endpoints a < b. Each point c ∈ [a, b] can be presented by the unique binomial convex combination INEQUALITIES ON THE LINE SEGMENT 49

c = αa + βb, (5) where

b − c c − a α = , β = . (6) b − a b − a

The next two lemmas present the properties of a convex function f : []a, b → R concerning its supporting and .

The discrete version refers to interval points and interval endpoints sharing the common center.

Lemma A. Let []a, b be a closed interval of real numbers, and let

n λ x be a convex combination of points x ∈ [a, b]. Let αa + βb be ∑i=1 i i i the unique endpoints convex combination such that

n ∑λixi = αa + βb. (7) i=1

Then every convex function f : [a, b] → R satisfies the double inequality

n f ()αa + βb ≤ ∑λi f ()()()xi ≤ αf a + βf b . (8) i=1

n Proof. Taking c = λ x , we have the following two cases. ∑i=1 i i

If c ∈ {}a, b , then Equation (8) is reduced to f (c) ≤ f (c) ≤ f (c).

If c ∈ ()a, b , then using a supporting line y = h1(x) of the convex curve y = f ()x at the graph point C(c, f (c)), and the secant line y = h2 ()x passing through the graph points A(a, f (a)) and B(b, f ()b ), we get the inequality 50 ZLATKO PAVIĆ

n f ()()αa + βb = h1 αa + βb = ∑λih1 ()xi i=1

n ≤ ∑λi f ()xi i=1

n ≤ ∑λih2 ()xi = h2 (αa + βb )= αf ()a + βf ()b , (9) i=1 containing Equation (8).  The discrete-integral version refers to the connection of the interval barycenter with interval endpoints.

Lemma B. Let []a, b be a closed interval of real numbers, and let µ be a positive measure on R such that µ([a, b]) > 0. Let αa + βb be the unique endpoints convex combination such that

1 x dµ = αa + βb. (10) µ()[]a, b ∫[]a,b

Then every convex function f : [a, b] → R satisfies the double inequality

1 f ()αa + βb ≤ f ()x dµ ≤ αf ()a + βf ()b . (11) µ()[]a, b ∫[]a,b

Proof. The proof can be done utilizing Equation (9) so that the integral means are used instead of the n-membered convex combinations.  We emphasize the basic content of Lemma A. Using the left-hand side of the inequality in Equation (8) with the n-membered convex combination instead of the binomial endpoints convex combination, we obtain the discrete form of Jensen’s inequality

 n  n f  λ x  ≤ λ f ()x . (12) ∑ i i  ∑ i i  i=1  i=1

Using the Riemann integral in Lemma B, the condition in (10) gives the midpoint INEQUALITIES ON THE LINE SEGMENT 51

1 b 1 1 x dx = a + b, (13) b − a ∫a 2 2 and its use in Equation (11) implies the classic Hermite-Hadamard inequality

 a + b  1 b f ()a + f ()b f   ≤ f ()x dx ≤ . (14)  2  b − a ∫a 2

Moreover, the inequality in Equation (14) follows by integrating the supporting-secant line inequality

h1(x) ≤ f (x) ≤ h2 (x), (15) over the interval []a, b .

We finish the section with a historic note on these two important inequalities. In 1905, applying the inductive principle, Jensen (see [4]) extended the inequality in Equation (2) to n-membered convex combinations. In 1906, working on transition to integrals, Jensen (see [5]) stated the another form. In 1883, studying convex functions, Hermite (see [3]) attained the inequality in Equation (14). In 1893, not knowing Hermite’s result, Hadamard (see [2]) got the left-hand side of Equation (14). For information as regards the Jensen and the Hermite-Hadamard inequalities, one may refer to papers [1], [6], [9], [10], [11], and [12].

3. Main Results

To refine the Hermite-Hadamard inequality in Equation (14), we will use convex combinations of points of the closed interval [a, b]. In the main Theorem 3.1, we improve Equation (14) by using convex combinations of the midpoint (a + b) 2. The concluding Theorem 3.4 presents the integral refinement of Equation (14).

We take points c, d ∈ [a, b] such that

c + d a + b = . (16) 2 2 52 ZLATKO PAVIĆ

Applying the right-hand side of the inequality in Equation (8) to the above assumption, and multiplying by 2, we obtain the simple inequality

f ()c + f (d) ≤ f (a) + f (b), (17) that will be used in this section. The main theorem follows.

Theorem 3.1. Let []a, b be a closed interval of real numbers, and let []c, d ⊂ ()a, b be a closed subinterval satisfying the common barycenter condition in Equation (16).

Then every convex function f : [a, b] → R satisfies the series of inequalities

 a + b  1  a + c  1  c + d  1  d + b  f   ≤ f   + f   + f    2  4  2  2  2  4  2 

c d b f ()x dx f ()x dx f ()x dx 1 ∫a 1 ∫c 1 ∫d ≤ + + 4 c − a 2 d − c 4 b − d

f ()a + 3f ()c + 3f (d) + f (b) f (a) + f (b) ≤ ≤ . (18) 8 2

Proof. Applying the Hermite-Hadamard inequality to the convex combination of points c and d written as

1 1 t = c + d, (19) 2 2 we have

 c + d  f ()t = f    2 

d f ()x dx ∫c ≤ d − c

f (c) + f (d) ≤ . (20) 2 INEQUALITIES ON THE LINE SEGMENT 53

Applying the same procedure to the convex combination of midpoints ()a + c 2 and ()d + b 2 given as

1 a + c 1 d + b t = + , (21) 2 2 2 2 we get

1  a + c  1  d + b  f ()t ≤ f   + f   2  2  2  2 

c b f ()x dx f ()x dx 1 ∫a 1 ∫d ≤ + 2 c − a 2 b − d

f (a) + f (c) + f (d) + f (b) ≤ . (22) 4

Taking the arithmetic means of the inequalities in Equations (20) and (22), using Equation (17) and rearranging, we obtain the inequality in Equation (18). 

The inequality in Equation (18) can be expressed using the point d = a + b − c. The observed Equation (18) can also be expressed with the point c = a + 2δ, where 0 < δ < (b − a) 2. Using this choice, we have d = b − 2δ, ()a + c 2 = a + δ and (d + b) 2 = b − δ. Finally, we can use the convex combinations

c = αa + βb, d = (1 − α)a + (1 − β)b, (23) provided that αa + βb < (a + b) 2. Regardless of all these cases, the midpoint ()a + b 2 is not covered on the right side of Equation (18).

The inequality in Equation (18) does not include the case c = d = ()a + b 2. A method similar to that in Theorem 3.1 can be applied to intervals []a, ()a + b 2 and [(a + b) 2, b] and so derive the inequality

 a + b  1   3a + b   a + 3b  f   ≤ f   + f    2  2   4   4  54 ZLATKO PAVIĆ

1 b ≤ f ()x dx b − a ∫a

1   a + b  f (a) + f (b) f (a) + f (b) ≤ f   + ≤ . (24) 2   2  2  2

The above improvement of the Hermite-Hadamard inequality was noted in [8].

A convex function f : [a, b] → R satisfies the inequality

f ()a + b − c ≤ f (a) + f (b) − f (c), (25) for every point c ∈ []a, b by Equation (17). The above simple inequality

n can be generalized by using the convex combination γ c instead of ∑i=1 i i the point c. Applying Jensen's inequality to the convex combination

n n t = a + b − ∑ γici = ∑ γi ()a + b − ci , (26) i=1 i=1 and using Equation (25), Mercer (see [7]) obtained the inequality

 n  n f a + b − γ c  ≤ f ()a + f ()b − γ f (c ). (27)  ∑ i i  ∑ i i  i=1  i=1

Corollary 3.2. Let []a, b be a closed interval of real numbers, let c ∈ ()a, b be an open interval point, and let f : [a, b] → R be a convex function.

If c ≤ ()a + b 2, then

c c f ()x dx + f ()x dx ∫a ∫a+b−c f ()()a + b − c + f c ≤ ≤ f ()a + f ()b . (28) c − a

If c ≥ ()a + b 2, then

a+b−c b f ()x dx + f ()x dx ∫a ∫c f ()()a + b − c + f c ≤ ≤ f ()a + f ()b . (29) c − a INEQUALITIES ON THE LINE SEGMENT 55

Proof. We prove the double inequality in Equation (28). Let y = h()x be the secant line of the convex curve y = f (x) passing through the graph points C()c, f ()c and D(a + b − c, f (a + b − c)). Using the affinity of the secant function h specified in Equations (1) and (4), and the inequality h(x)()≤ f x for x ∈ [a, c] ∪ [a + b − c, b], we get

1 1 1 1 f ()c + f ()a + b − c = h()c + h()a + b − c 2 2 2 2

 c b   x dx + x dx   a + b   ∫a ∫a+b−c  = h  = h  2   2()c − a     

c b c b h()x dx + h ()x dx f ()x dx + f ()x dx ∫a ∫a+b−c ∫a ∫a+b−c = ≤ . (30) 2()c − a 2()c − a

Continuing with Equation (30) by applying the Hermite-Hadamard inequality and the inequality in Equation (25), it follows

f ()a + f ()c + f (a + b − c) + f (b) f (a) + f (b) ≤ ≤ . (31) 4 2

It remains only to multiply by 2. 

Corollary 3.3. Let []a, b be a closed interval of real numbers, and let []c, d ⊂ ()a, b be a closed subinterval satisfying the common barycenter condition in Equation (16).

Then every convex function f : [a, b] → R satisfies the double integral inequality

d b c b f ()x dx f ()x dx f ()x dx f ()x dx ∫c ∫a ∫a ∫d ≤ ≤ + . (32) d − c b − a 2()c − a 2()b − d 56 ZLATKO PAVIĆ

Proof. Midpoint equality in Equation (16) can be expressed in the integral form by the barycenter equalities

d b c b x dx x dx x dx x dx ∫c ∫a ∫a ∫d = = + . (33) d − c b − a 2()c − a 2()b − d

We firstly prove that the left term of Equation (32) is less than or equal to the right term. Let y = h(x) be the secant line of the convex curve y = f ()x passing through the graph points C(c, f (c)) and D(d, f ()d ). Using inequalities f ()x ≤ h(x) for x ∈ [c, d], and h(x)()≤ f x for x ∈ [][a, c ∪ a + b − c, d ], and applying the affinity of function h to Equation (33), we get

d d c b f ()x dx h()x dx h()x dx h()x dx ∫c ∫c ∫a ∫d ≤ = + d − c d − c 2()c − a 2()b − d

c b f ()x dx f ()x dx ∫a ∫d ≤ + . (34) 2()c − a 2()b − d

Now, the inequality in Equation (32) can be confirmed by the combination

b c d b f ()x dx f ()x dx + f ()x dx + f ()x dx ∫a ∫a ∫c ∫d = b − a b − a

 d   c b   f ()x dx   f ()x dx f ()x dx  ∫c ∫a ∫d = α   + β  +  , (35)  d − c   2()c − a 2()b − d          which is convex because the coefficients

d − c 2(c − a) 2(d − b) α = , β = = (36) b − a b − a b − a are nonnegative and their sum is equal to 1.  INEQUALITIES ON THE LINE SEGMENT 57

Combining the inequalities in Equations (14), (32), (28), and (29), we get the following integral refinement of the Hermite-Hadamard inequality.

Theorem 3.4. Let []a, b be a closed interval of real numbers, and let []c, d ⊂ ()a, b be a closed subinterval satisfying the common barycenter condition in Equation (16). Then every convex function f : [a, b] → R satisfies the series of inequalities

d b f ()x dx f ()x dx  a + b  ∫c ∫a f   ≤ ≤  2  d − c b − a

c b f ()x dx f ()x dx ∫a ∫d f ()a + f ()b ≤ + ≤ . (37) 2()c − a 2()b − d 2

4. Application to Means

Thorough this section we use positive numbers a and b, and a strictly monotone continuous function ϕ : [a, b] → R.

The discrete quasi-arithmetic mean of the numbers a and b respecting the function ϕ can be defined by the number

−1 1 1  M ()a, b = ϕ  ϕ()a + ϕ()b . (38) ϕ  2 2 

Using the identity function ϕ(x) = x, we get the generalized arithmetic mean

1 1 A()a, b = a + b, (39) 2 2 using the hyperbolic function ϕ(x) = 1 x, we have the generalized

−1  1 −1 1 −1  H()a, b =  a + b  , (40)  2 2  58 ZLATKO PAVIĆ and using the logarithmic function ϕ(x) = ln x, we obtain the generalized geometric mean

1 1 G()a, b = a 2b 2 . (41)

The above means satisfy the generalized harmonic-geometric-arithmetic mean inequality

H()a, b < G(a, b) < A(a, b). (42)

The integral quasi-arithmetic mean of the numbers a and b respecting the function ϕ can be defined by the number

b −1 1  M ϕ ()a, b = ϕ  ϕ()x dx . (43)  b − a ∫a 

Using the hyperbolic function, we have the logarithmic mean

−1  1 b 1  b − a L()a, b =  dx  = , (44)  b − a ∫a x  ln b − ln a and using the logarithmic function, we obtain the identric mean

1  1 b  1  bb  b−a I()a, b = exp  ln x dx  =   . (45)    a   b − a ∫a  e  a 

The well-known mean inequality asserts that

H()a, b < G ()a, b < L(a, b) < I(a, b) < A(a, b). (46)

Applying Equation (18) to the convex function f (x) = − ln x using substitutions a 1 a, b 1 b, c 1 c, and d 1 d, and then acting on the rearranged inequality with the , we can derive the series of inequalities

INEQUALITIES ON THE LINE SEGMENT 59

H()()a, c H c, d H(d, b) H()a, b ≤ 32

I −1()()()a−1, c−1 I −1 c−1, d−1 I −1 d−1, b−1 ≤ 32

≤ ()ac3d3b 1 8 ≤ G()a, b , (47) refining the harmonic-geometric mean inequality.

Applying Equation (37) to the exponential function f ()x = ex using substitutions a ln a, b ln b, c ln c, and d ln d, we obtain the series of inequalities

G()a, b ≤ L ()c, d ≤ L(a, b)

1 1 ≤ L()a, c + L()d, b ≤ A ()a, b , (48) 2 2 refining the geometric-logarithmic-arithmetic mean inequality.

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